Question

The altitude of a triangle is increasing at a rate of 2.0 centimeters/minute while the area of the triangle is increasing at a rate of 4.0 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 10.5 centimeters and the area is 84.0 square centimeters?

Answer 1

i guess this was not clear the first time, so lets go slow. you know
\[A'(t)=40, h'(t)=2\] you also know \[A=\frac{1}{2}bh\] that is
Area is one half base times height. you want
\[b'(t)\]

Answer 2

excuse me i meant

\[A'(t)=4\] not 40

Answer 3

take the derivative of the formula
\[A=\frac{1}{2}bh\] wrt time to get
\[A'(t)=\frac{1}{2}(b'h + bh')\] via product rule

Answer 4

we can solve for h' by
\[2A'=b'h+bh'\]
\[2A'-b'h=bh'\]
\[h'=\frac{2A'-b'h}{b}\]

Answer 5

got it, -2.2857 is my answer. thanks for your help :]

Answer 6

yw.